3.169 \(\int \frac{x^2 (a+b \sinh ^{-1}(c x))}{(d+c^2 d x^2)^{5/2}} \, dx\)

Optimal. Leaf size=119 \[ \frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 d \left (c^2 d x^2+d\right )^{3/2}}-\frac{b}{6 c^3 d^2 \sqrt{c^2 x^2+1} \sqrt{c^2 d x^2+d}}-\frac{b \sqrt{c^2 x^2+1} \log \left (c^2 x^2+1\right )}{6 c^3 d^2 \sqrt{c^2 d x^2+d}} \]

[Out]

-b/(6*c^3*d^2*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2]) + (x^3*(a + b*ArcSinh[c*x]))/(3*d*(d + c^2*d*x^2)^(3/2))
- (b*Sqrt[1 + c^2*x^2]*Log[1 + c^2*x^2])/(6*c^3*d^2*Sqrt[d + c^2*d*x^2])

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Rubi [A]  time = 0.124617, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {5723, 266, 43} \[ \frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 d \left (c^2 d x^2+d\right )^{3/2}}-\frac{b}{6 c^3 d^2 \sqrt{c^2 x^2+1} \sqrt{c^2 d x^2+d}}-\frac{b \sqrt{c^2 x^2+1} \log \left (c^2 x^2+1\right )}{6 c^3 d^2 \sqrt{c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2)^(5/2),x]

[Out]

-b/(6*c^3*d^2*Sqrt[1 + c^2*x^2]*Sqrt[d + c^2*d*x^2]) + (x^3*(a + b*ArcSinh[c*x]))/(3*d*(d + c^2*d*x^2)^(3/2))
- (b*Sqrt[1 + c^2*x^2]*Log[1 + c^2*x^2])/(6*c^3*d^2*Sqrt[d + c^2*d*x^2])

Rule 5723

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
((f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n)/(d*f*(m + 1)), x] - Dist[(b*c*n*d^IntPart[p]*(d + e
*x^2)^FracPart[p])/(f*(m + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*Arc
Sinh[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && EqQ[m + 2*p
+ 3, 0] && NeQ[m, -1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\left (d+c^2 d x^2\right )^{5/2}} \, dx &=\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 d \left (d+c^2 d x^2\right )^{3/2}}-\frac{\left (b c \sqrt{1+c^2 x^2}\right ) \int \frac{x^3}{\left (1+c^2 x^2\right )^2} \, dx}{3 d^2 \sqrt{d+c^2 d x^2}}\\ &=\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 d \left (d+c^2 d x^2\right )^{3/2}}-\frac{\left (b c \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{x}{\left (1+c^2 x\right )^2} \, dx,x,x^2\right )}{6 d^2 \sqrt{d+c^2 d x^2}}\\ &=\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 d \left (d+c^2 d x^2\right )^{3/2}}-\frac{\left (b c \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \left (-\frac{1}{c^2 \left (1+c^2 x\right )^2}+\frac{1}{c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )}{6 d^2 \sqrt{d+c^2 d x^2}}\\ &=-\frac{b}{6 c^3 d^2 \sqrt{1+c^2 x^2} \sqrt{d+c^2 d x^2}}+\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 d \left (d+c^2 d x^2\right )^{3/2}}-\frac{b \sqrt{1+c^2 x^2} \log \left (1+c^2 x^2\right )}{6 c^3 d^2 \sqrt{d+c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.196037, size = 118, normalized size = 0.99 \[ -\frac{\sqrt{c^2 d x^2+d} \left (-2 a c^3 x^3 \sqrt{c^2 x^2+1}+b c^2 x^2+b \left (c^2 x^2+1\right )^2 \log \left (c^2 x^2+1\right )-2 b c^3 x^3 \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)+b\right )}{6 c^3 d^3 \left (c^2 x^2+1\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(a + b*ArcSinh[c*x]))/(d + c^2*d*x^2)^(5/2),x]

[Out]

-(Sqrt[d + c^2*d*x^2]*(b + b*c^2*x^2 - 2*a*c^3*x^3*Sqrt[1 + c^2*x^2] - 2*b*c^3*x^3*Sqrt[1 + c^2*x^2]*ArcSinh[c
*x] + b*(1 + c^2*x^2)^2*Log[1 + c^2*x^2]))/(6*c^3*d^3*(1 + c^2*x^2)^(5/2))

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Maple [B]  time = 0.161, size = 1153, normalized size = 9.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(5/2),x)

[Out]

-1/3*a/c^2*x/d/(c^2*d*x^2+d)^(3/2)+1/3*a/c^2/d^2*x/(c^2*d*x^2+d)^(1/2)+2/3*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)
^(1/2)/c^3/d^3*arcsinh(c*x)+b*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)*c^4/d^3*arcsi
nh(c*x)*x^7-b*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)*c^3/d^3*arcsinh(c*x)*(c^2*x^2
+1)^(1/2)*x^6+1/6*b*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)*c^4/d^3*x^7-1/6*b*(d*(c
^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)*c^2/d^3*(c^2*x^2+1)*x^5+b*(d*(c^2*x^2+1))^(1/2)/
(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)*c^2/d^3*arcsinh(c*x)*x^5-2*b*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c
^6*x^6+10*c^4*x^4+5*c^2*x^2+1)*c/d^3*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x^4+1/3*b*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8
+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)*c^2/d^3*x^5-1/2*b*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c
^2*x^2+1)*c/d^3*(c^2*x^2+1)^(1/2)*x^4-1/6*b*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)
/d^3*(c^2*x^2+1)*x^3+1/3*b*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)/d^3*arcsinh(c*x)
*x^3-4/3*b*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)/c/d^3*arcsinh(c*x)*(c^2*x^2+1)^(
1/2)*x^2+1/6*b*(d*(c^2*x^2+1))^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)/d^3*x^3-1/2*b*(d*(c^2*x^2+1)
)^(1/2)/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)/c/d^3*x^2*(c^2*x^2+1)^(1/2)-1/3*b*(d*(c^2*x^2+1))^(1/2)/(
3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)/c^3/d^3*arcsinh(c*x)*(c^2*x^2+1)^(1/2)-1/6*b*(d*(c^2*x^2+1))^(1/2)
/(3*c^8*x^8+9*c^6*x^6+10*c^4*x^4+5*c^2*x^2+1)/c^3/d^3*(c^2*x^2+1)^(1/2)-1/3*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1
)^(1/2)/c^3/d^3*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)

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Maxima [A]  time = 1.25048, size = 185, normalized size = 1.55 \begin{align*} -\frac{1}{6} \, b c{\left (\frac{1}{c^{6} d^{\frac{5}{2}} x^{2} + c^{4} d^{\frac{5}{2}}} + \frac{\log \left (c^{2} x^{2} + 1\right )}{c^{4} d^{\frac{5}{2}}}\right )} + \frac{1}{3} \, b{\left (\frac{x}{\sqrt{c^{2} d x^{2} + d} c^{2} d^{2}} - \frac{x}{{\left (c^{2} d x^{2} + d\right )}^{\frac{3}{2}} c^{2} d}\right )} \operatorname{arsinh}\left (c x\right ) + \frac{1}{3} \, a{\left (\frac{x}{\sqrt{c^{2} d x^{2} + d} c^{2} d^{2}} - \frac{x}{{\left (c^{2} d x^{2} + d\right )}^{\frac{3}{2}} c^{2} d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

-1/6*b*c*(1/(c^6*d^(5/2)*x^2 + c^4*d^(5/2)) + log(c^2*x^2 + 1)/(c^4*d^(5/2))) + 1/3*b*(x/(sqrt(c^2*d*x^2 + d)*
c^2*d^2) - x/((c^2*d*x^2 + d)^(3/2)*c^2*d))*arcsinh(c*x) + 1/3*a*(x/(sqrt(c^2*d*x^2 + d)*c^2*d^2) - x/((c^2*d*
x^2 + d)^(3/2)*c^2*d))

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c^{2} d x^{2} + d}{\left (b x^{2} \operatorname{arsinh}\left (c x\right ) + a x^{2}\right )}}{c^{6} d^{3} x^{6} + 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} + d^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(c^2*d*x^2 + d)*(b*x^2*arcsinh(c*x) + a*x^2)/(c^6*d^3*x^6 + 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 + d^3),
x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*asinh(c*x))/(c**2*d*x**2+d)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )} x^{2}}{{\left (c^{2} d x^{2} + d\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arcsinh(c*x))/(c^2*d*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)*x^2/(c^2*d*x^2 + d)^(5/2), x)